A 2.3 kg pendulum bob is pulled out such that its center of mass is 0.26 m above

Question

A 2.3 kg pendulum bob is pulled out such that its center of mass is 0.26 m above its lowest point and released. At the lowest point in its swing it strikes a 1.5 kg block initially at rest on a frictionless surface.

a) What is the gravitational potential energy of the pendulum initially.

b) What is the speed of the pendulum bob just before striking the block?

c) If the pendulum bob continues forward with a speed of 1.1 m/s, what is the speed of the block after the collision?

d) Is the collision elastic? Explain your answer.

Explanation / Answer

a) PE = M g h = 2.3 * 9.8 * .26 = 5.86 J
b) PE = KE = 1/2 M V1^2 = 5.86
V1 = (2 * 5.86 / 2.3)^1/2 = 2.26 m/s     speed of bob before collision
c) M V1 = M V2 + m Vb        conservation of momentum
Vb = M (V1 - V2) / m = 2.3 * (2.26 - 1.1) / 1.5 = 1.78 m/s
d) KE2 = 1/2 M V2^2 + 1/2 m Vb^2       total KE after collision
KE2 = 1/2 * 2.3 * 1.1^2 + 1/2 * 1.5 * 1.78^2 = 3.77 J
Energy lost in collision = 5.86 - 3.77 = 2.09 J
so the collision is inelastic because energy is not conserved

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